Fixing $nge 1$, consider the permutation $p$ on $n$ elements whose word representation consists of the even integers ascending up to $n$, followed by the odd integers descending from $n$. Let $q$ be the reverse of $p$. The permutations are perhaps more easily defined via example: when $n=8$, we have $p=24687531$ and $q=13578642$.
Here’s a natural question: What is the group $G$ generated by the permutations $p,q$?
Using GAP, I have computed these groups for $1le nle 128$. The answer is almost a straightforward one: $A_n$ when $n=0,1pmod{4}$, and $S_n$ when $n=2,3pmod{4}$. That is, it will be as large as possible, given the parity of $p$ and $q$.
However, this classification has the following exceptions:

When $n=2^k$ for $kge2$, $G$ is a combination of direct and semidirect products of certain cyclic groups, with an order of $2^kcdot (k+1)$; I think it is always given by the semidirect product $(C_2)^krtimes C_{k+1}$, though this doesn’t always match how GAP’s
StructureDescription()
function interprets the group. (When $k=2$, this coincides with the expected $A_4$.) 
When $n=6$, $G$ is not $S_6$, but $S_5$.

When $n=12$, $G$ is not $A_{12}$, but the sporadic simple group $M_{12}$.
I have two questions about these groups and their apparent classification.

Can we understand how the $n=6$ and $n=12$ sporadic cases work? I suspect at least the $n=6$ case is related somehow to the sporadic outer automorphisms of $S_6$, but I am not quite sure how this connection manifests.

Can the classification outlined above be proven for all other $n$, if indeed it is true? (If not, can a counterexample be found?)
For convenience, the Sage code used to reach the above conclusions is linked here, and can be run here.